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\includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-3_597_951_1471_598} In the diagram the tangent to a curve at a general point \(P\) with coordinates \(( x , y )\) meets the \(x\)-axis at \(T\). The point \(N\) on the \(x\)-axis is such that \(P N\) is perpendicular to the \(x\)-axis. The curve is such that, for all values of \(x\) in the interval \(0 < x < \frac { 1 } { 2 } \pi\), the area of triangle \(P T N\) is equal to \(\tan x\), where \(x\) is in radians.

1. Using the fact that the gradient of the curve at \(P\) is \(\frac { P N } { T N }\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 } y ^ { 2 } \cot x .$$
2. Given that \(y = 2\) when \(x = \frac { 1 } { 6 } \pi\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).